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3106. Lexicographically Smallest String After Operations With Constraint

Description

You are given a string s and an integer k.

Define a function distance(s1, s2) between two strings s1 and s2 of the same length n as:

  • The sum of the minimum distance between s1[i] and s2[i] when the characters from 'a' to 'z' are placed in a cyclic order, for all i in the range [0, n - 1].

For example, distance("ab", "cd") == 4, and distance("a", "z") == 1.

You can change any letter of s to any other lowercase English letter, any number of times.

Return a string denoting the lexicographically smallest string t you can get after some changes, such that distance(s, t) <= k.

 

Example 1:

Input: s = "zbbz", k = 3

Output: "aaaz"

Explanation:

Change s to "aaaz". The distance between "zbbz" and "aaaz" is equal to k = 3.

Example 2:

Input: s = "xaxcd", k = 4

Output: "aawcd"

Explanation:

The distance between "xaxcd" and "aawcd" is equal to k = 4.

Example 3:

Input: s = "lol", k = 0

Output: "lol"

Explanation:

It's impossible to change any character as k = 0.

 

Constraints:

  • 1 <= s.length <= 100
  • 0 <= k <= 2000
  • s consists only of lowercase English letters.

Solutions

Solution 1: Enumeration

We can traverse each position of the string $s$. For each position, we enumerate all characters less than the current character, calculate the cost $d$ to change to this character. If $d \leq k$, we change the current character to this character, subtract $d$ from $k$, end the enumeration, and continue to the next position.

After the traversal, we get a string that meets the conditions.

The time complexity is $O(n \times |\Sigma|)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$, and $|\Sigma|$ is the size of the character set. In this problem, $|\Sigma| \leq 26$.

  • class Solution {
    public String getSmallestString(String s, int k) {
        char[] cs = s.toCharArray();
        for (int i = 0; i < cs.length; ++i) {
            char c1 = cs[i];
            for (char c2 = 'a'; c2 < c1; ++c2) {
                int d = Math.min(c1 - c2, 26 - c1 + c2);
                if (d <= k) {
                    cs[i] = c2;
                    k -= d;
                    break;
                }
            }
        }
        return new String(cs);
    }
}

  • class Solution {
public:
    string getSmallestString(string s, int k) {
        for (int i = 0; i < s.size(); ++i) {
            char c1 = s[i];
            for (char c2 = 'a'; c2 < c1; ++c2) {
                int d = min(c1 - c2, 26 - c1 + c2);
                if (d <= k) {
                    s[i] = c2;
                    k -= d;
                    break;
                }
            }
        }
        return s;
    }
};

  • class Solution:
    def getSmallestString(self, s: str, k: int) -> str:
        cs = list(s)
        for i, c1 in enumerate(s):
            for c2 in ascii_lowercase:
                if c2 >= c1:
                    break
                d = min(ord(c1) - ord(c2), 26 - ord(c1) + ord(c2))
                if d <= k:
                    cs[i] = c2
                    k -= d
                    break
        return "".join(cs)


  • func getSmallestString(s string, k int) string {
	cs := []byte(s)
	for i, c1 := range cs {
		for c2 := byte('a'); c2 < c1; c2++ {
			d := int(min(c1-c2, 26-c1+c2))
			if d <= k {
				cs[i] = c2
				k -= d
				break
			}
		}
	}
	return string(cs)
}

  • function getSmallestString(s: string, k: number): string {
    const cs: string[] = s.split('');
    for (let i = 0; i < s.length; ++i) {
        for (let j = 97; j < s[i].charCodeAt(0); ++j) {
            const d = Math.min(s[i].charCodeAt(0) - j, 26 - s[i].charCodeAt(0) + j);
            if (d <= k) {
                cs[i] = String.fromCharCode(j);
                k -= d;
                break;
            }
        }
    }
    return cs.join('');
}

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